¹⁰⁰/₆ = 16, remainder = 4. Hence 2 more should be added to 100 to get the minimum
3 digit number divisible by 6.
=> Minimum 3 digit number divisible by 6 = 100 + 2 = 102
⁹⁹⁹/₆ = 166, remainder = 3. Hence 3 should be decreased from 999 to get the maximum
3 digit number divisible by 6.
=> Maximum 3 digit number divisible by 6 = 999 - 3 = 996
Hence, the 3 digit numbers divisible by 6 are 102, 108, 114,... 996
This is Arithmetic Progression with a = 102 ,d = 6, l=996

$$\text{Number of terms }= \dfrac{(l-a)}{d} + 1 = \dfrac{(996-102)}{6} + 1 = \dfrac{894}{6} + 1 = 149 + 1 = 150 $$